docs: def K_branch_converging_Crane(D_run, D_branch, Q_run, Q_branch, angle=90):
      r'''Returns the loss coefficient for the branch of a converging tee or wye
      according to the Crane method [1]_.
      
      .. math::
          K_{branch} = C\left[1 + D\left(\frac{Q_{branch}}{Q_{comb}\cdot 
          \beta_{branch}^2}\right)^2 - E\left(1 - \frac{Q_{branch}}{Q_{comb}}
          \right)^2 - \frac{F}{\beta_{branch}^2} \left(\frac{Q_{branch}}
          {Q_{comb}}\right)^2\right]
              
      .. math::
          \beta_{branch} = \frac{D_{branch}}{D_{comb}}
      
      In the above equation, D = 1, E = 2. See the notes for definitions of F and
      C.
  
      Parameters
      ----------
      D_run : float
          Diameter of the straight-through inlet portion of the tee or wye [m]
      D_branch : float
          Diameter of the pipe attached at an angle to the straight-through, [m]
      Q_run : float
          Volumetric flow rate in the straight-through inlet of the tee or wye,
          [m^3/s]
      Q_branch : float
          Volumetric flow rate in the pipe attached at an angle to the straight-
          through, [m^3/s]
      angle : float, optional
          Angle the branch makes with the straight-through (tee=90, wye<90)
          [degrees]
  
      Returns
      -------
      K : float
          Loss coefficient of branch with respect to the velocity and inside 
          diameter of the combined flow outlet [-]
  
      Notes
      -----
      F is linearly interpolated from the table of angles below. There is no 
      cutoff to prevent angles from being larger or smaller than 30 or 90
      degrees. 
  
      +-----------+------+
      | Angle [°] |      |
      +===========+======+
      | 30        | 1.74 |
      +-----------+------+
      | 45        | 1.41 |
      +-----------+------+
      | 60        | 1    |
      +-----------+------+
      | 90        | 0    |
      +-----------+------+
      
      If :math:`\beta_{branch}^2 \le 0.35`, C = 1
  
      If :math:`\beta_{branch}^2 > 0.35` and :math:`Q_{branch}/Q_{comb} > 0.4`,
      C = 0.55.
  
      If neither of the above conditions are met:
          
      .. math::
          C = 0.9\left(1 - \frac{Q_{branch}}{Q_{comb}}\right)
  
      Note that there is an error in the text of [1]_; the errata can be obtained 
      here: http://www.flowoffluids.com/publications/tp-410-errata.aspx
  
      Examples
      --------
      Example 7-35 of [1]_. A DN100 schedule 40 tee has 1135 liters/minute of
      water passing through the straight leg, and 380 liters/minute of water
      converging with it through a 90° branch. Calculate the loss coefficient in
      the branch. The calculated value there is -0.04026.
      
      >>> K_branch_converging_Crane(0.1023, 0.1023, 0.018917, 0.00633)
      -0.04044108513625682
      
      References
      ----------
      .. [1] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane,
         2009.
      '''
      beta = (D_branch/D_run)
      beta2 = beta*beta
      Q_comb = Q_run + Q_branch
      Q_ratio = Q_branch/Q_comb
      if beta2 <= 0.35:
          C = 1.
      elif Q_ratio <= 0.4:
          C = 0.9*(1 - Q_ratio)
      else:
          C = 0.55
      D, E = 1., 2.
      F = interp(angle, branch_converging_Crane_angles, branch_converging_Crane_Fs)
      K = C*(1. + D*(Q_ratio/beta2)**2 - E*(1. - Q_ratio)**2 - F/beta2*Q_ratio**2)
      return K